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3 years ago

A 0.450-kg soccer ball has a kinetic energy of 119 J. What is the velocity of the soccer ball?

Physics

1 answer:

Elis [28]3 years ago

3 0

Answer: 23m/s

Explanation:

Given that:

Mass of soccer ball = 0.450kg

kinetic energy = 119 J

velocity of the soccer ball = ?

Kinetic energy is the energy possessed by a moving object. It is measured in joules, and depends on the mass (m) of the object and the velocity (v) by which it moves

i.e K.E = 1/2mv^2

119J = 1/2 x 0.450kg x v^2

119J = 0.225kg x v^2

v^2 = 119J / 0.225kg

v^2 = 528.9

To get the value of velocity, find the square root of v

v = √528.9

v = 22.997m/s (Round result to 23)

Thus, the velocity of the soccer ball is 23m/s

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

3 years ago

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I  is 1360 W/m²

P sun = 2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

6 0

3 years ago

An atom has 5 protons, 6 neutrons, and 5 electrons. what is the atomic mass?

ozzi

Answer:

the atomic mass is 11

Explanation:

the atomic mass is basically how many protons and neutrons there are so for this all you have to do is some simple math:

5 + 6 = 11

and boom, ur atomic mass is equal to 11!

4 0

2 years ago

What is the best structure for a egg dropping project you will be name brainiest

julsineya [31]

Answer:

bubble wrap in stuff animal

Explanation:

did it

4 0

3 years ago

Read 2 more answers

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = $\sqrt{2gh}$

V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

= 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

= - 0.329 J

Hence, kinetic energy is not conserved.

8 0

3 years ago